The line 2y+x=10 meets the circle x^2+y^2=65 at P and Q. Calculate the length of PQ?

Hi Tae, if we were to draw both of these on a graph we would get a circle centered at the origin and a line that meets the circle at two points.
So, to find the answer, we need to find the values of x and y for which the equations are equal. This turns into a simultaneous equation. When we find the final values, these will represent the two points (P,Q) that we then find the distance between.
For clarity, I want to rewrite 2y + x = 10 as y = 5 - x/2.
A: x^2 + y^2 = 65
B: y = 5 - x/2
Here is my general solution:
x^2 + (5 - x/2)^2 = 65 (insert B into A)
x^2 + 25 - 10x/2 + x^2/4 = 65 (expanded (5 - x/2)^2)
(5x^2)/4 - 5x - 40 = 0 (simplify)
5x^2 - 20x - 160 = 0 (simplify more to get a quadratic equation, woo!)
(5x - 40)(x + 4) = 0 (factorise)
x = -4 and x = 8 (find possible solutions to equation)
y = 7 and y = 1 (insert values of x from above into equation B)
(-4, 7) and (8, 1) (now we have two co-ordinate sets)
((8 - (-4))^2 + (1 - 7)^2)^(1/2) (use the formula ((x2 - x1)^2 - (y2 - y1)^2)^(1/2) to find the distance between two points on a straight line)
180^(1/2) (simplify)
There we have it Tae! I hope you find that helped :)